You need to understand: 1) do-notation 2) Monad instance for List 1) do-notation is just syntactic sugar around (>>=) Monad operator. So: filterM p (x:xs) = p x >>= \flg -> filterM p xs >>= \ys -> if flg then x:ys else ys 2) Monad instance for List: http://hackage.haskell.org/package/base-4.8.0.0/docs/src/GHC-Base.html#line-... In particular: xs >>= f = [y | x <- xs, y <- f x] or the equivalent: xs >>= f = concatMap f xs -- filterM specialized for [] filterM :: (a -> [Bool]) -> [a] -> [[a]] filterM _ [] = [[]] filterM p (x:xs) = concatMap (\flg -> concatMap (\ys -> [if flg then x:ys else ys]) (filterM p xs)) (p x) powerset :: [a] -> [[a]] powerset xs = filterM (\x -> [True, False]) xs i.e. powerset [] = [[]] powerset (x:xs) = concatMap (\flg -> concatMap (\ys -> [if flg then x:ys else ys]) (powerset xs)) [True,False] i.e. powerset [] = [[]] powerset (x:xs) = concatMap (\flg -> fmap (\ys -> if flg then x:ys else ys) (powerset xs)) [True,False] i.e. powerset [] = [[]] powerset (x:xs) = let ys = powerset xs in fmap (x:) ys ++ ys I would directly use the latter form instead of using filterM to implement powerset. Sylvain 2015-04-22 11:22 GMT+02:00 Shishir Srivastava :

Hi Mike,

Thanks for your response. I was aware that 'casting' doesn't really fit in Haskell vocab but for the lack of better word used it.

My question was however more towards the usage of [Bool] in the 'if' statement of the filterM function.

More precisely - How does 'if [True, False] then x else y' work , because when I do this in GHCi it throws up the following error ?

...

...

Couldn't match expected type `Bool' with actual type `[Bool]'

Clearly the 'if' construct does not take a list of Boolean but a single Boolean value so how does filterM use it in it's implementation.

Hope have made myself clear this time.

Thanks, Shishir

...

From: Mike Meyer To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Cc: Date: Wed, 22 Apr 2015 04:15:31 -0500 Subject: Re: [Haskell-beginners] filterM function

On Wed, Apr 22, 2015 at 3:31 AM, Shishir Srivastava wrote:

...

I still don't quite understand how 'flg' being a boolean [] is used in the last 'if statement' of implementation because when I try to do the same thing outside in GHCi it fails miserably even though I am casting it to [Int] -

-- return (if [True, False] then "4" else "3")::[Int]

"cast" is a misnomer in Haskell. When you add a type to an expression, you aren't changing the type of the expression like a C-style cast, but picking out a type from the set of possible types for that expression.

Ignoring the if part and and focusing on return, which has a type Monad m => a -> m a. [Int] is equivalent to [] Int, so [] would be the Monad, and a is Int. While 3 can be an Int, "3", can't. So you could do return 3 :: [Int] or equivalently return 3 :: [] Int to get [3], you can't do return "3" :: [Int], because "3" can't be an Int. You can do return "3" :: [String], since "3" is a string. Just to show the range of possibilities, you can do return 3 :: IO Float, and get back 3.0 as an IO action. The monad in the type is IO, and 3 can be interpreted as a Float.

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